SOLUTION: CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 145 seconds (2 minutes and 25 seconds). This will allow the disk jockeys to have plenty of time for commercials with

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Question 1091883: CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 145 seconds (2 minutes and 25 seconds). This will allow the disk jockeys to have plenty of time for commercials within each 10-minute segment. Assume the distribution of the length of the cuts follows the normal distribution with a population standard deviation of 10 seconds. Suppose we select a sample of 24 cuts from various CDs sold by CRA CDs Inc.

a. What can we say about the shape of the distribution of the sample mean?

Sample mean

b. What is the standard error of the mean? (Round your z-value to 2 decimal places and final answer to 2 decimal places.)

Standard error of the mean
seconds.

c.
What percent of the sample means will be greater than 150 seconds? (Round your z-value to 2 decimal places and final answer to 2 decimal places.)

Percent
%

d.
What percent of the sample means will be greater than 138 seconds? (Round your answer to 2 decimal places.)

Percent
%

e.
What percent of the sample means will be greater than 138 but less than 150 seconds? (Round your z-value to 2 decimal places and final answer to 2 decimal places.)

Percent
%

Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
a. The sample mean is normally distributed with a smaller standard deviation and the same mean as the population, 2m25s
b. The standard error of the mean is sigma/sqrt (n)=10 sec/sqrt(24)=2.04 sec. The std error is the std deviation of the sample. There is no z-value involved here.
c. z > 2.45(xbar-mean)/SE=(150-145)/2.04. This would be 0.71% or 0.0071
d. z > (138-145)/2.04=-7/2.04 or z > -3.43 or 99.97% or 0.9997
d. z>150 sec is 0.0071 from above, so we would subtract that from d to get 0.9926 or 99.26%.